package main

/**
请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。
[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。
示例 1：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：
输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false
提示：
1 <= board.length <= 200
1 <= board[i].length <= 200
*/
func exist(board [][]byte, word string) bool {
	words := []byte(word)
	for i := 0; i < len(board); i++ {
		for j := 0; j < len(board[0]); j++ {
			if dfs(board, words, i, j, 0) {
				return true
			}
		}
	}
	return false
}

func dfs(board [][]byte, words []byte, i int, j int, k int) bool {
	if i < 0 || i >= len(board) || j < 0 || j >= len(board[0]) || board[i][j] != words[k] {
		return false
	}
	if k == len(words)-1 {
		return true
	}
	board[i][j] = '/'
	b := dfs(board, words, i-1, j, k+1) || dfs(board, words, i+1, j, k+1) || dfs(board, words, i, j-1, k+1) || dfs(board, words, i, j+1, k+1)
	board[i][j] = words[k]
	return b
}

func main() {

}
